K - 4.8=0.32 - 5.4k answer?

Mathematics

1 answer:

RSB [31]2 years ago

5 0

Answer:

k =4/5

Step-by-step explanation:

$k - 4.8=0.32 - 5.4k\\\mathrm{Multiply\:both\:sides\:by\:}100\\k\times\:100-4.8\times\:100=0.32\times\:100-5.4k\times\:100\\\\Refine\\100k-480=32-540k\\\\Add\: 480\mathrm{\:to\:both\:sides}\\100k-480+480=32-540k+480\\\\Simplify\\100k=-540k+512\\\\\mathrm{Add\:}540k\mathrm{\:to\:both\:sides}\\100k+540k=-540k+512+540k\\\\Simplify\\640k=512\\\\\mathrm{Divide\:both\:sides\:by\:}640\\\frac{640k}{640}=\frac{512}{640}\\\\k=\frac{4}{5}$

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HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$